3cosA=5sinA then find the value of 5sinA-2sec^3A+2cosA
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Answered by
1
Answer:
......
........
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Answered by
0
Step-by-step explanation:
3sinA+5cosA=5
⇒(3sinA+5cosA)
2
=25
⇒9sin
2
A+25cos
2
A+30sinAcosA=25 ……(1)
Let (3cosA−5sinA)
2
=k
⇒9cos
2
A+25sin
2
A−30sinAcosA=k …….(2)
(1)+(2)
⇒9(sin
2
A+cos
2
A)+25(cos
2
A+sin
2
A)=25+k
⇒9+25=25+k
⇒k=9.
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